Finite length transmission line

Discussion in 'General Electronics Chat' started by atomtm, Jul 27, 2014.

  1. atomtm

    Thread Starter New Member

    Aug 13, 2012
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    Hello everyone !
    I am reading about transmission lines and standing waves in them , because of impedance difference.
    The concept of an open circuit and it's reflections is perfectly clear , though I can not grasp the situation where the end of the line is short-circuited .
    As the text - book says :
    " A similar phenomenon takes place if the end of a transmission line is short-circuited: when the voltage wave-front reaches the end of the line, it is reflected back to the source, because voltage cannot exist between two electrically common points. When this reflected wave reaches the source, the source sees the entire transmission line as a short-circuit. Again, this happens as quickly as the signal can propagate round-trip down and up the transmission line at whatever velocity allowed by the dielectric material between the line's conductors. ".

    This is very confusing and if anyone could explain this more ( with an example like the train in the textbook ) I would be very grateful.
    Thank you in advance
     
  2. crutschow

    Expert

    Mar 14, 2008
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    The energy in the pulse is stored in the capacitance and inductance of the transmission line. When the line is short-circuited the pulse energy goes through the short. The pulse energy has to go somewhere so it goes back to the transmission but now is stored as a negative pulse travelling back down the line (since the energy goes from a positive pulse center-to-shield, to a positive pulse, shield-to-center). If, at that time the pulse source is still high, the negative pulse will cancel the positive pulse, giving 0V (so it looks like a short). If the pulse source is low by that time, then you will see the negative pulse.

    Make sense?
     
    Last edited: Jul 27, 2014
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  3. GopherT

    AAC Fanatic!

    Nov 23, 2012
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  4. atomtm

    Thread Starter New Member

    Aug 13, 2012
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    Can you explain this a bit more ?
    GopherT thank you for the video I will check this later out , dont have accesss to youtube right now

    Also , is there any analogy like the train example given in textbook for the short circuit ?

    Thank you for your help !
    Really appreciate it
     
  5. alfacliff

    Well-Known Member

    Dec 13, 2013
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    and to make it even more complicated, the effect reverses every quarter wave. a shorted odd quarterwavelength acts like an open, even quarterwavelengths act like shorts.
     
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  6. atomtm

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    Aug 13, 2012
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    nice!!!!:d
     
  7. crutschow

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    Mar 14, 2008
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    Assume a positive pulse (with respect to the outside shield) is going down the center conductor. When it reaches the short at the end, the pulse travels directly to the shield. This means the pulse is now positive at the shield with respect to the center (or negative from the center conductor to the shield).

    Since I don't know what train analogy you are referring to, I can't comment on that.
     
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  8. atomtm

    Thread Starter New Member

    Aug 13, 2012
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    By the train example I mean the example provided in the textbook (Vol 2 , AC , Transmission lines ) :

    If the end of a transmission line is open-circuited -- that is, left unconnected -- the current “wave” propagating down the line's length will have to stop at the end, since electrons cannot flow where there is no continuing path. This abrupt cessation of current at the line's end causes a “pile-up” to occur along the length of the transmission line, as the electrons successively find no place to go. Imagine a train traveling down the track with slack between the rail car couplings: if the lead car suddenly crashes into an immovable barricade, it will come to a stop, causing the one behind it to come to a stop as soon as the first coupling slack is taken up, which causes the next rail car to stop as soon as the next coupling's slack is taken up, and so on until the last rail car stops. The train does not come to a halt together, but rather in sequence from first car to last: (Figure below)
    [​IMG]

    A similar phenomenon takes place if the end of a transmission line is short-circuited: when the voltage wave-front reaches the end of the line, it is reflected back to the source, because voltage cannot exist between two electrically common points. When this reflected wave reaches the source, the source sees the entire transmission line as a short-circuit. Again, this happens as quickly as the signal can propagate round-trip down and up the transmission line at whatever velocity allowed by the dielectric material between the line's conductors.

    Thank you for your time!
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    The wave travels down the line and part of it gets reflected whenever there is a change in impedance, small or large. An open circuit or a short circuit is a large change, while the two conductors pressed a little closer together near (say) the center is a small change, but even that small change causes a reflection.

    The reflection happens because of the conservation of energy, and energy can come in the form of a larger voltage and smaller current, or a larger current and a smaller voltage, or anywhere in between.

    Viewing the energy as small packets like train cars, as the energy is handed off down the line eventually it comes to the end where it can only be handed off back up the line, but the line still has energy being handed off forward as well as the additional energy being handed off backward. This means that the energy at any point along the line will be caused by the sum of the forward signal and the backward signal. For sinusoidal waves this means there will be one sinusoid pair V(x,t), I(x,t) going down the line, and one pair going back up the line, and their vector sum makes up the total response. This creates humps where they add and dips where they subtract.
     
    Last edited: Jul 28, 2014
  10. atomtm

    Thread Starter New Member

    Aug 13, 2012
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